96.106 Problem number 181

\[ \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^3} \, dx \]

Optimal antiderivative \[ -\frac {1}{2 \left (b x -\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )\right ) \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}+\frac {1}{\left (b x -\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )\right )^{2} \mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )}-\frac {\ln \left (x \right )}{\left (b x -\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )\right )^{3}}+\frac {\ln \left (\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )\right )}{\left (b x -\mathrm {arccoth}\left (\tanh \left (b x +a \right )\right )\right )^{3}} \]

command

integrate(1/x/arccoth(tanh(b*x+a))^3,x, algorithm="giac")

Giac 1.9.0-11 via sagemath 9.6 output

\[ \frac {4 \, {\left (-3 i \, \pi - 4 \, b x - 6 \, a\right )}}{4 \, \pi ^{2} b^{2} x^{2} - 16 i \, \pi a b^{2} x^{2} - 16 \, a^{2} b^{2} x^{2} + 4 i \, \pi ^{3} b x + 24 \, \pi ^{2} a b x - 48 i \, \pi a^{2} b x - 32 \, a^{3} b x - \pi ^{4} + 8 i \, \pi ^{3} a + 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} - 16 \, a^{4}} - \frac {8 i \, \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{3} - 6 i \, \pi ^{2} a - 12 \, \pi a^{2} + 8 i \, a^{3}} + \frac {8 i \, \log \left (x\right )}{\pi ^{3} - 6 i \, \pi ^{2} a - 12 \, \pi a^{2} + 8 i \, a^{3}} \]

Giac 1.7.0 via sagemath 9.3 output

\[ \int \frac {1}{x \operatorname {arcoth}\left (\tanh \left (b x + a\right )\right )^{3}}\,{d x} \]________________________________________________________________________________________